Physics Q2 – NEET-UG 2025

A microscope has an objective of focal length

2 cm

, eyepiece of focal length

4 cm

and the tube length of

40 cm

. If the distance of distinct vision of eye is

25 cm

, the magnification in the microscope is

Clue 1 of 3: Compound Microscope Magnification

The total magnification of a compound microscope is the product of the magnification due to the objective lens and the magnification due to the eyepiece.

🧠 Full Solution Path

Step 1: Identify Given Parameters✦ Active

The given parameters are:

- Focal length of objective, $f_{o} = 2 cm$

- Focal length of eyepiece, $f_{e} = 4 cm$

- Tube length, $L = 40 cm$

- Distance of distinct vision, $D = 25 cm$

Step 2: Apply Magnification Formula for Normal Adjustment○ Expand

For a compound microscope, when the final image is formed at infinity (normal adjustment), the total magnification $M$ is given by the formula:

M = \frac{L}{f_{o}} \times \frac{D}{f_{e}}

This formula is often used in competitive exams when an exact option matches this interpretation, even if the question mentions the distance of distinct vision as a parameter.

💡 Teacher's Secret Hint

Be aware that 'tube length' can sometimes refer to the distance between the focal points, or the distance between the lenses. Also, the final image can be at infinity or at the distance of distinct vision. Choose the interpretation that leads to an exact match with the options if ambiguity exists.

Step 3: Calculate Total Magnification○ Expand

Substitute the given values into the formula:

M = \frac{40 cm}{2 cm} \times \frac{25 cm}{4 cm}

M = 20 \times 6.25

M = 125

💡 Teacher's Secret Hint

Ensure all units are consistent before calculation. In this case, all lengths are in centimeters, so no conversion is needed.

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Physics Question 2 – NEET-UG 2025

Clue 1 of 3: Compound Microscope Magnification

🧠 Full Solution Path